Re: [MS_AccessPros] SQLFMM example: SQL not matching Design Grid in Access

 

Jo-

Which edition of the book are you working from?  In the latest edition, the problem statement is:

 5. "Show me customers and employees who have the same last name."

(Hint: The solution requires a JOIN on matching values.)

You can find the solution in CH08_Customers_Employees_Same_LastName

(16 rows).

The SQL for the query is:

SELECT (Customers.CustFirstName & ' ' & Customers.CustLastName) AS CustFullName, (Employees.EmpFirstName & ' ' & Employees.EmpLastName) AS EmpFullName

FROM Customers INNER JOIN Employees ON Customers.CustLastName=Employees.EmpLastName;

.. which matches what you have in your message.

The output is:

CustFullName	EmpFullName
Rachel Patterson	Ann Patterson
Luke Patterson	Ann Patterson
Maria Patterson	Ann Patterson
Neil Patterson	Ann Patterson
William Thompson	Mary Thompson
Suzanne Viescas	Carol Viescas
John Viescas	Carol Viescas
Caleb Viescas	Carol Viescas
Kirk DeGrasse	Kirk DeGrasse
Suzanne Viescas	David Viescas
John Viescas	David Viescas
Caleb Viescas	David Viescas
Rachel Patterson	Kathryn Patterson
Luke Patterson	Kathryn Patterson
Maria Patterson	Kathryn Patterson
Neil Patterson	Kathryn Patterson

16 rows.  What are you seeing different?

John Viescas, Author

Effective SQL

SQL Queries for Mere Mortals 

Microsoft Access 2010 Inside Out

Microsoft Access 2007 Inside Out

Microsoft Access 2003 Inside Out

Building Microsoft Access Applications 

http://www.viescas.com/ 

(Paris, France)

On Dec 19, 2016, at 1:46 AM, jovaughn@rochester.rr.com [MS_Access_Professionals] <MS_Access_Professionals@yahoogroups.com> wrote:

I am working thru SQLFMM. In CH08 example the SQL output does not match what shows up and is required 

in the Design Grid. Is this a well known feature/bug?

SELECT (Customers.CustFirstName & ' ' & Customers.CustLastName) AS CustFullName, 

            (Employees.EmpFirstName & ' ' & Employees.EmpLastName) AS EmpFullName

FROM Customers INNER JOIN Employees 

            ON Customers.CustLastName = Employees.EmpLastName;

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Posted by: John Viescas <johnv@msn.com>

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